In this MathOverflow post on visualizing high-dimensional spaces, Terry Tao states that "the fact that most of the mass of a unit ball in high dimensions lurks near the boundary of the ball can be interpreted as a manifestation of the law of large numbers, using the interpretation of a high-dimensional vector space as the state space for a large number of trials of a random variable."

What exactly does he mean?


The only interpretation I've come up with is the following:
Observing n trials of a random variable (for instance $X \sim U(0,1)$) is equivalent to picking a random point inside an n-dimensional unit cube. For large n, most of the volume of the n-dimensional unit cube is near the edges. Thus our point is likely to be near an edge of the cube, which means one of our trials has a large value. This shows some law of rare events: "given enough trials, a rare event will occur".

However, this interpretation must use an n-cube rather than an n-sphere because the trials are independent. Additionally, our conclusion is not the law of large numbers, which states that the average of our trials will converge to the expected value for large n.

Can someone provide a correct interpretation of Tao's comment?


  • $\begingroup$ "For large n, most of the volume of the n-dimensional unit cube is near the edges". Why do you say this? The mass density of unit cube equals to 1 and does not depend on dimensionality. $\endgroup$ – Sasha May 22 '12 at 0:28
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    $\begingroup$ My reasoning was: Consider a point "near an edge" if it is in the unit cube (s=1) but not in a cube of s=0.9. Now the proportion of volume "near an edge" is 1^n - (0.9)^n which approaches 1 as n gets large. $\endgroup$ – tba May 22 '12 at 0:36
  • $\begingroup$ For a precise version of the hypercube intuition, see here. The vertices of the hypercube can be viewed as realizations of sequences of i.i.d. Bernoulli(1/2) random variables. We know that the binomial distribution peaks when the number of 1's and 0's is approximately equal. $\endgroup$ – sai May 29 '12 at 4:01

One possibility is to generate points in the $n$-ball by taking a sample size $n+1$ from $X \sim N(0,1)$, let $\lambda = \sqrt{\sum_{i=1}^{n+1}X_i^2}$ and looking at $(X_1/\lambda,X_2/\lambda, \ldots , X_n/\lambda)$ as random points in the ball.

The square of the distance from the centre, namely ${\sum_{i=1}^{n}(X_i/\lambda)^2}$, will be more concentrated close $1$ as $n$ increases.

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    $\begingroup$ Nice interpretation, but since there is no sample average going on - this does not seem to be directly related to the LLN, no? $\endgroup$ – sai May 29 '12 at 4:16
  • $\begingroup$ Agreed, I spent some time trying to understand this but couldn't find the probabilistic interpretation of (X1/l, X2/l, ..., Xn/l). Is there a better interpretation than "A sample size n normalized by the stddev of n+1 samples"? $\endgroup$ – tba May 30 '12 at 4:46

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